Suppose we have a function defined as follows:
$\alpha(f,x_o)=\limsup\{|f(a)-f(b)|:a,b\in (x_o-\frac{1}{n},x_o+\frac{1}{n})\}$
with $f:R\rightarrow R$ and $x_o \in R$. I need to prove that $\alpha=0$ iff $f$ is continuous at $x_o$.
It seems trivial to show the $\alpha=0 \Rightarrow$ direction, since $|f(a)-f(b)|\geq0$, so if the $lim\,sup$ is $0$ we must have $f(a)=f(b)$ everywhere on the interval.
However, I'm really drawing a blank on the other direction. How can I prove that continuity of $f$ implies that $\alpha=0$? It seems like I should use the $\epsilon-\delta$ definition of continuity, since I'm working with $|f(a)-f(b)|$, but I think the $lim\,sup$ stuff is throwing me off the scent. Any help would be much appreciated!
Let $\epsilon >0$. If $n$ is sufficiently large then (by continuity) $|f(a)-f(b)| \leq |f(a)-f(x_0)|+|f(x_0)-f(b)|<2\epsilon$ whenever $a,b\in (x_0-\frac 1 n, x_0+\frac 1 n)$. Take sup over all such $a,b$ and then let $n \to \infty$. You will get $\alpha (f,x_0) \leq 2\epsilon$. Since this is true for all $\epsilon >0$ we must have $\alpha (f,x_0)=0$.